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Banach fixed-point theorem
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The well known fixed-point theorem by Banach reads as follows:
Let $(X,d)$ be a complete metric space, and $Asubseteq X$ closed. Let $f: Ato A$ be a function, and $gamma$ a constant with $0leqgamma <1$, such that $d(f(x), f(y))leqgammacdot d(x,y)$ for every $x,yin A$.
Define $(x_n)_ninmathbbN$ by $x_n+1= f(x_n)$ for an arbitrary starting point $x_0in A$.
Then exists exactly one $ain A$ such that $f(a)=a$ and $lim_ntoinfty x_n=a$ for an arbitrary $x_0$.
Obviously you can look up a proof easily, but I have a theorem, which states the following, and I am curious if you can use it to proof Banach's fixed-point theorem.
Let $(X,d)$ be a complete metric space. Let $A_0supset A_1supsetdotso$ be a decreasing sequence of nonempty closed subsets with $operatornamediam(A_n)stackrelntoinftylongrightarrow 0$, where $operatornamediam(A_n):=supd(x,y)$.
Then exists exactly one unique point in $bigcap_ninmathbbN A_n$
The theorems are kinda familiar and my guess is, that you can construct such a decreasing sequence of sets, such that $bigcap_ninmathbbN A_n=a$. Where $a$ is the desired fixed point.
If it is possible to use this efficiently, I would appreciate a hint, to try it myself.
Thanks in advance.
fixed-point-theorems
asked Jul 19 at 20:02
Cornman
2,47921127
add a comment |Â
up vote
3
down vote
favorite
The well known fixed-point theorem by Banach reads as follows:
Let $(X,d)$ be a complete metric space, and $Asubseteq X$ closed. Let $f: Ato A$ be a function, and $gamma$ a constant with $0leqgamma <1$, such that $d(f(x), f(y))leqgammacdot d(x,y)$ for every $x,yin A$.
Define $(x_n)_ninmathbbN$ by $x_n+1= f(x_n)$ for an arbitrary starting point $x_0in A$.
Then exists exactly one $ain A$ such that $f(a)=a$ and $lim_ntoinfty x_n=a$ for an arbitrary $x_0$.
Obviously you can look up a proof easily, but I have a theorem, which states the following, and I am curious if you can use it to proof Banach's fixed-point theorem.
Let $(X,d)$ be a complete metric space. Let $A_0supset A_1supsetdotso$ be a decreasing sequence of nonempty closed subsets with $operatornamediam(A_n)stackrelntoinftylongrightarrow 0$, where $operatornamediam(A_n):=supd(x,y)$.
Then exists exactly one unique point in $bigcap_ninmathbbN A_n$
The theorems are kinda familiar and my guess is, that you can construct such a decreasing sequence of sets, such that $bigcap_ninmathbbN A_n=a$. Where $a$ is the desired fixed point.
If it is possible to use this efficiently, I would appreciate a hint, to try it myself.
Thanks in advance.
fixed-point-theorems
asked Jul 19 at 20:02
Cornman
2,47921127
Choose $A_0 = A$ and $A_n+1 := overlinef(A_n)$.
– amsmath
Jul 19 at 20:10
@amsmath For example if $A=Bbb R$ and $f(x)=(x+1)/2$...
– David C. Ullrich
Jul 19 at 22:57
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The well known fixed-point theorem by Banach reads as follows:
Let $(X,d)$ be a complete metric space, and $Asubseteq X$ closed. Let $f: Ato A$ be a function, and $gamma$ a constant with $0leqgamma <1$, such that $d(f(x), f(y))leqgammacdot d(x,y)$ for every $x,yin A$.
Define $(x_n)_ninmathbbN$ by $x_n+1= f(x_n)$ for an arbitrary starting point $x_0in A$.
Then exists exactly one $ain A$ such that $f(a)=a$ and $lim_ntoinfty x_n=a$ for an arbitrary $x_0$.
Obviously you can look up a proof easily, but I have a theorem, which states the following, and I am curious if you can use it to proof Banach's fixed-point theorem.
Let $(X,d)$ be a complete metric space. Let $A_0supset A_1supsetdotso$ be a decreasing sequence of nonempty closed subsets with $operatornamediam(A_n)stackrelntoinftylongrightarrow 0$, where $operatornamediam(A_n):=supd(x,y)$.
Then exists exactly one unique point in $bigcap_ninmathbbN A_n$
The theorems are kinda familiar and my guess is, that you can construct such a decreasing sequence of sets, such that $bigcap_ninmathbbN A_n=a$. Where $a$ is the desired fixed point.
If it is possible to use this efficiently, I would appreciate a hint, to try it myself.
Thanks in advance.
fixed-point-theorems
asked Jul 19 at 20:02
Cornman
2,47921127
The well known fixed-point theorem by Banach reads as follows:
Let $(X,d)$ be a complete metric space, and $Asubseteq X$ closed. Let $f: Ato A$ be a function, and $gamma$ a constant with $0leqgamma <1$, such that $d(f(x), f(y))leqgammacdot d(x,y)$ for every $x,yin A$.
Define $(x_n)_ninmathbbN$ by $x_n+1= f(x_n)$ for an arbitrary starting point $x_0in A$.
Then exists exactly one $ain A$ such that $f(a)=a$ and $lim_ntoinfty x_n=a$ for an arbitrary $x_0$.
Obviously you can look up a proof easily, but I have a theorem, which states the following, and I am curious if you can use it to proof Banach's fixed-point theorem.
Let $(X,d)$ be a complete metric space. Let $A_0supset A_1supsetdotso$ be a decreasing sequence of nonempty closed subsets with $operatornamediam(A_n)stackrelntoinftylongrightarrow 0$, where $operatornamediam(A_n):=supd(x,y)$.
Then exists exactly one unique point in $bigcap_ninmathbbN A_n$
The theorems are kinda familiar and my guess is, that you can construct such a decreasing sequence of sets, such that $bigcap_ninmathbbN A_n=a$. Where $a$ is the desired fixed point.
If it is possible to use this efficiently, I would appreciate a hint, to try it myself.
Thanks in advance.
fixed-point-theorems
asked Jul 19 at 20:02
Cornman
2,47921127
asked Jul 19 at 20:02
Cornman
2,47921127
asked Jul 19 at 20:02
Cornman
2,47921127
asked Jul 19 at 20:02
Cornman
2,47921127
2,47921127
Choose $A_0 = A$ and $A_n+1 := overlinef(A_n)$.
– amsmath
Jul 19 at 20:10
@amsmath For example if $A=Bbb R$ and $f(x)=(x+1)/2$...
– David C. Ullrich
Jul 19 at 22:57
add a comment |Â
Choose $A_0 = A$ and $A_n+1 := overlinef(A_n)$.
– amsmath
Jul 19 at 20:10
@amsmath For example if $A=Bbb R$ and $f(x)=(x+1)/2$...
– David C. Ullrich
Jul 19 at 22:57
Choose $A_0 = A$ and $A_n+1 := overlinef(A_n)$.
– amsmath
Jul 19 at 20:10
Choose $A_0 = A$ and $A_n+1 := overlinef(A_n)$.
– amsmath
Jul 19 at 20:10
@amsmath For example if $A=Bbb R$ and $f(x)=(x+1)/2$...
– David C. Ullrich
Jul 19 at 22:57
@amsmath For example if $A=Bbb R$ and $f(x)=(x+1)/2$...
– David C. Ullrich
Jul 19 at 22:57
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The suggestion $A_0=A$, $A_n+1=overlinef(A_n)$ only works if $A$ has finite diameter.
You could derive the first theorem from the second by setting $$A_n=overlinex_n,x_n+1,dots.$$
I'm not suggesting that's the right way to prove Banach's theorem! It's certainly not the "efficient" argument you asked for, since working out the details seems like more work then just proving Banach's theorem by the method you know. But it does serve to show how the two results are related, even if $A$ has infinite diameter.
answered Jul 19 at 22:56
David C. Ullrich
54.2k33582
add a comment |Â
up vote
1
down vote
Hint: Let $A_0 = A$ and $A_n+1 = mathrmcl(f"A_n)$ (the closure of the pointwise image of $A_n$ under $f$).
answered Jul 19 at 20:11
Stefan Mesken
13.6k32045
What do you mean by "pointwise image of $A_n$ under $f$"? Simply, that you look at the image $f(A_n)$ and then take the closure? It is a mix of a lack of english skills and confusion by the notation $f ''$.
– Cornman
Jul 19 at 20:14
1
@Cornman Yes. $f" A_n = f(x) mid x in A_n$ -- call that set $B_n$. And then I take $A_n+1$ to be the closure of $B_n$ in your metric space. (In set theory, we don't like to write $f(A_n)$, since that can lead to confusion in case that $A_n$ is both a subset and an element of the domain of $f$ -- that's why we write $f" A_n$ for the pointwise image instead).
– Stefan Mesken
Jul 19 at 20:16
1
That works if the diameter of $A$ is finite...
– David C. Ullrich
Jul 19 at 21:48
@DavidC.Ullrich Good point.
– Stefan Mesken
Jul 19 at 21:50
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The suggestion $A_0=A$, $A_n+1=overlinef(A_n)$ only works if $A$ has finite diameter.
You could derive the first theorem from the second by setting $$A_n=overlinex_n,x_n+1,dots.$$
I'm not suggesting that's the right way to prove Banach's theorem! It's certainly not the "efficient" argument you asked for, since working out the details seems like more work then just proving Banach's theorem by the method you know. But it does serve to show how the two results are related, even if $A$ has infinite diameter.
answered Jul 19 at 22:56
David C. Ullrich
54.2k33582
add a comment |Â
up vote
1
down vote
accepted
The suggestion $A_0=A$, $A_n+1=overlinef(A_n)$ only works if $A$ has finite diameter.
You could derive the first theorem from the second by setting $$A_n=overlinex_n,x_n+1,dots.$$
I'm not suggesting that's the right way to prove Banach's theorem! It's certainly not the "efficient" argument you asked for, since working out the details seems like more work then just proving Banach's theorem by the method you know. But it does serve to show how the two results are related, even if $A$ has infinite diameter.
answered Jul 19 at 22:56
David C. Ullrich
54.2k33582
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The suggestion $A_0=A$, $A_n+1=overlinef(A_n)$ only works if $A$ has finite diameter.
You could derive the first theorem from the second by setting $$A_n=overlinex_n,x_n+1,dots.$$
I'm not suggesting that's the right way to prove Banach's theorem! It's certainly not the "efficient" argument you asked for, since working out the details seems like more work then just proving Banach's theorem by the method you know. But it does serve to show how the two results are related, even if $A$ has infinite diameter.
answered Jul 19 at 22:56
David C. Ullrich
54.2k33582
The suggestion $A_0=A$, $A_n+1=overlinef(A_n)$ only works if $A$ has finite diameter.
You could derive the first theorem from the second by setting $$A_n=overlinex_n,x_n+1,dots.$$
I'm not suggesting that's the right way to prove Banach's theorem! It's certainly not the "efficient" argument you asked for, since working out the details seems like more work then just proving Banach's theorem by the method you know. But it does serve to show how the two results are related, even if $A$ has infinite diameter.
answered Jul 19 at 22:56
David C. Ullrich
54.2k33582
answered Jul 19 at 22:56
David C. Ullrich
54.2k33582
answered Jul 19 at 22:56
David C. Ullrich
54.2k33582
answered Jul 19 at 22:56
David C. Ullrich
54.2k33582
54.2k33582
add a comment |Â
add a comment |Â
up vote
1
down vote
Hint: Let $A_0 = A$ and $A_n+1 = mathrmcl(f"A_n)$ (the closure of the pointwise image of $A_n$ under $f$).
answered Jul 19 at 20:11
Stefan Mesken
13.6k32045
What do you mean by "pointwise image of $A_n$ under $f$"? Simply, that you look at the image $f(A_n)$ and then take the closure? It is a mix of a lack of english skills and confusion by the notation $f ''$.
– Cornman
Jul 19 at 20:14
1
@Cornman Yes. $f" A_n = f(x) mid x in A_n$ -- call that set $B_n$. And then I take $A_n+1$ to be the closure of $B_n$ in your metric space. (In set theory, we don't like to write $f(A_n)$, since that can lead to confusion in case that $A_n$ is both a subset and an element of the domain of $f$ -- that's why we write $f" A_n$ for the pointwise image instead).
– Stefan Mesken
Jul 19 at 20:16
1
That works if the diameter of $A$ is finite...
– David C. Ullrich
Jul 19 at 21:48
@DavidC.Ullrich Good point.
– Stefan Mesken
Jul 19 at 21:50
add a comment |Â
up vote
1
down vote
Hint: Let $A_0 = A$ and $A_n+1 = mathrmcl(f"A_n)$ (the closure of the pointwise image of $A_n$ under $f$).
answered Jul 19 at 20:11
Stefan Mesken
13.6k32045
What do you mean by "pointwise image of $A_n$ under $f$"? Simply, that you look at the image $f(A_n)$ and then take the closure? It is a mix of a lack of english skills and confusion by the notation $f ''$.
– Cornman
Jul 19 at 20:14
1
@Cornman Yes. $f" A_n = f(x) mid x in A_n$ -- call that set $B_n$. And then I take $A_n+1$ to be the closure of $B_n$ in your metric space. (In set theory, we don't like to write $f(A_n)$, since that can lead to confusion in case that $A_n$ is both a subset and an element of the domain of $f$ -- that's why we write $f" A_n$ for the pointwise image instead).
– Stefan Mesken
Jul 19 at 20:16
1
That works if the diameter of $A$ is finite...
– David C. Ullrich
Jul 19 at 21:48
@DavidC.Ullrich Good point.
– Stefan Mesken
Jul 19 at 21:50
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Let $A_0 = A$ and $A_n+1 = mathrmcl(f"A_n)$ (the closure of the pointwise image of $A_n$ under $f$).
answered Jul 19 at 20:11
Stefan Mesken
13.6k32045
Hint: Let $A_0 = A$ and $A_n+1 = mathrmcl(f"A_n)$ (the closure of the pointwise image of $A_n$ under $f$).
answered Jul 19 at 20:11
Stefan Mesken
13.6k32045
answered Jul 19 at 20:11
Stefan Mesken
13.6k32045
answered Jul 19 at 20:11
Stefan Mesken
13.6k32045
answered Jul 19 at 20:11
Stefan Mesken
13.6k32045
13.6k32045
What do you mean by "pointwise image of $A_n$ under $f$"? Simply, that you look at the image $f(A_n)$ and then take the closure? It is a mix of a lack of english skills and confusion by the notation $f ''$.
– Cornman
Jul 19 at 20:14
1
@Cornman Yes. $f" A_n = f(x) mid x in A_n$ -- call that set $B_n$. And then I take $A_n+1$ to be the closure of $B_n$ in your metric space. (In set theory, we don't like to write $f(A_n)$, since that can lead to confusion in case that $A_n$ is both a subset and an element of the domain of $f$ -- that's why we write $f" A_n$ for the pointwise image instead).
– Stefan Mesken
Jul 19 at 20:16
1
That works if the diameter of $A$ is finite...
– David C. Ullrich
Jul 19 at 21:48
@DavidC.Ullrich Good point.
– Stefan Mesken
Jul 19 at 21:50
add a comment |Â
What do you mean by "pointwise image of $A_n$ under $f$"? Simply, that you look at the image $f(A_n)$ and then take the closure? It is a mix of a lack of english skills and confusion by the notation $f ''$.
– Cornman
Jul 19 at 20:14
1
@Cornman Yes. $f" A_n = f(x) mid x in A_n$ -- call that set $B_n$. And then I take $A_n+1$ to be the closure of $B_n$ in your metric space. (In set theory, we don't like to write $f(A_n)$, since that can lead to confusion in case that $A_n$ is both a subset and an element of the domain of $f$ -- that's why we write $f" A_n$ for the pointwise image instead).
– Stefan Mesken
Jul 19 at 20:16
1
That works if the diameter of $A$ is finite...
– David C. Ullrich
Jul 19 at 21:48
@DavidC.Ullrich Good point.
– Stefan Mesken
Jul 19 at 21:50
What do you mean by "pointwise image of $A_n$ under $f$"? Simply, that you look at the image $f(A_n)$ and then take the closure? It is a mix of a lack of english skills and confusion by the notation $f ''$.
– Cornman
Jul 19 at 20:14
What do you mean by "pointwise image of $A_n$ under $f$"? Simply, that you look at the image $f(A_n)$ and then take the closure? It is a mix of a lack of english skills and confusion by the notation $f ''$.
– Cornman
Jul 19 at 20:14
1
1
@Cornman Yes. $f" A_n = f(x) mid x in A_n$ -- call that set $B_n$. And then I take $A_n+1$ to be the closure of $B_n$ in your metric space. (In set theory, we don't like to write $f(A_n)$, since that can lead to confusion in case that $A_n$ is both a subset and an element of the domain of $f$ -- that's why we write $f" A_n$ for the pointwise image instead).
– Stefan Mesken
Jul 19 at 20:16
@Cornman Yes. $f" A_n = f(x) mid x in A_n$ -- call that set $B_n$. And then I take $A_n+1$ to be the closure of $B_n$ in your metric space. (In set theory, we don't like to write $f(A_n)$, since that can lead to confusion in case that $A_n$ is both a subset and an element of the domain of $f$ -- that's why we write $f" A_n$ for the pointwise image instead).
– Stefan Mesken
Jul 19 at 20:16
1
1
That works if the diameter of $A$ is finite...
– David C. Ullrich
Jul 19 at 21:48
That works if the diameter of $A$ is finite...
– David C. Ullrich
Jul 19 at 21:48
@DavidC.Ullrich Good point.
– Stefan Mesken
Jul 19 at 21:50
@DavidC.Ullrich Good point.
– Stefan Mesken
Jul 19 at 21:50
add a comment |Â
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Choose $A_0 = A$ and $A_n+1 := overlinef(A_n)$.
– amsmath
Jul 19 at 20:10
@amsmath For example if $A=Bbb R$ and $f(x)=(x+1)/2$...
– David C. Ullrich
Jul 19 at 22:57